Question 1002742
P = orginal amount of carbon 14 t years ago (right when the person died)
A = current amount of carbon 14


Since "An anthropologist finds bone that her instruments measure it as 0.177% of the amount of Carbon-14 the bones would have contained when the person was alive" we know that A = 0.00177P because 0.177% = 0.00177



The half-life formula in this case is A = P*(1/2)^(t/5730)



A = P*(1/2)^(t/5730)



0.00177P = P*(1/2)^(t/5730) <font color=blue>Plug in A = 0.00177P</font>



0.00177 = (1/2)^(t/5730) <font color=blue>Divide both sides by P</font>



Ln(0.00177) = Ln((1/2)^(t/5730)) <font color=blue>Apply natural logs to both sides</font>



Ln(0.00177) = (t/5730)*Ln(1/2) <font color=blue>Use the rule Ln(x^y) = y*Ln(x)</font>



Ln(0.00177)/Ln(1/2) = t/5730 <font color=blue>Divide both sides by Ln(1/2)</font>



5730*Ln(0.00177)/Ln(1/2) = t <font color=blue>Multiply both sides by 5730</font>



t = 5730*Ln(0.00177)/Ln(1/2) <font color=blue>Flip the equation (a = b is the same as b = a)</font>



t = 52,383.8601165473 <font color=blue>Use a calculator</font>



t = 52,383.860  <font color=blue> Rounding to the nearest thousand (3 decimal places)</font>



The person died approximately <font size = 5 color=red>52,383.860</font> years ago